I read that in the Wikipedia page. There is some obvious argument involving repeatedly switching sides and whatnot, and it has some subtle logical flaw. Why bother with that at all.

OK, but let's look at it:

1. Denote by A the amount in the player's selected envelope.

2. The probability that A is the smaller amount is 1/2, and that it is the larger amount is also 1/2.

3. The other envelope may contain either 2A or A/2.

OK, here is where this argument got seduced by the multiplicative thinking. The correct reasoning is: let's call the smaller base amount A and the larger one 2A. One envelope contains A, the other one 2A. That's it.

3. If A is the smaller amount, then the other envelope contains 2A.

4. If A is the larger amount, then the other envelope contains A/2.

Here, the argument is splitting into cases because of the IF, and in these two cases, A has a different meaning.

5. Thus the other envelope contains 2A with probability 1/2 and A/2 with probability 1/2.

6. So the expected value of the money in the other envelope is:

This is wonky. The overall two-envelope situation has an expected value, not the individual envelope.

The correct view is that if this is the richer envelope, it contains X more money. If it is the less endowed one, then it contains X less.

    1        1           5
    - 2A +   - (A/2) =   - A
    2        2           4 

7. This is greater than A so, on average, the person reasons that they stand to gain by swapping.

That calculation is wrong though, by itself. That issue should be attacked directly.

Separate calculations have to be performed for the two cases: A is the smaller amount and A is the bigger amount. For each of these cases, the expected value from both envelopes has to be calculated.

The way the above formula is doing the expected value calculation is off because it's equivocating on A.

It ends up working with two unrelated quantities: 2A and A/2. 2A is four times bigger than A/2. A is simultaneously denoting the smaller amount and the larger one.

8. After the switch, denote that content by B and reason in exactly the same manner as above.

Yes, and if you do, you will end up with 5/4B. So that can't be right and the whole remaining argument about infinite switching being irrational compared to just opening an envelope is superfluous.

Because the variable A refers to an amount, has to have a consistent meaning across the two cases, and so it has to be:

1. We have A; the other envelope has 2A; or else

2. We have 2A; and the other envelope has A.

In case 1, if we switch we gain A.

In case 2, if we switch, we lose A.

We have no idea which case we are in.

We cannot use the variable A to denote our initially selected envelope in both cases, because then A simultaneously represents the smaller amount and the larger amount, and we are absurdly mixing the quantities A/2 and 2A in the same calculation. No quantities in the actual problem are related by a factor of four!

People might be seduced into the A equivocation perhaps because they are initially thinking of A as being a label for the envelope, which is arbitrary and innocent enough. We have some envelope A and it has some money in it. Oh, but the other envelope then has either A/2 or 2A. At that point, A is no longer an envelope name, but a name for an amount. Then it has to pin down a specific number: it has to denote either the smaller value, or the larger one, and that value is found in different envelopes in the different cases; A cannot both be "name of initially chosen envelope" and "amount in that envelope".